# Algebraically closed group

Algebraically closed group

In mathematics, in the realm of group theory, a group $A$ is algebraically closed if any finite set of equations and inequations that "make sense" in $A$ already have a solution in $A$. This idea will be made precise later in the article.

Informal discussion

Suppose we wished to find an element $x$ of a group $G$ satisfying the conditions (equations and inequations):

::$x^2=1$::$x^3=1$::$x e 1$

Then it is easy to see that this is impossible because the first two equations imply $x=1$. In this case we say the set of conditions are inconsistent with $G$. (In fact this set of conditions are inconsistent with any group whatsoever.)

However if we extend the group $G$ to the group $H$ with multiplication table:

Then the condions have two solutions, namely $x=b$ and $x=c$.

Thus there are three possibilities regarding such conditions:
* They may be inconsistent with $G$ and have no solution in any extension of $G$.
* They may have a solution in $G$.
* They may have no solution in $G$ but nevertheless have a solution in some extension $H$ of $G$.

It is reasonable to ask whether there are any groups $A$ such that whenever a set of conditions like these have a solution at all, they have a solution in $A$ itself? The answer turns out to be "yes", and we call such groups algebraically closed groups.

Formal definition of an algebraically closed group

We first need some preliminary ideas.

If $G$ is a group and $F$ is the free group on countably many generators, then by a finite set of equations and inequations with coefficients in $G$ we mean a pair of subsets $E$ and $I$ of $Fstar G$ the free product of $F$ and $G$.

This formalizes the notion of a set of equations and inequations consisting of variables $x_i$ and elements $g_j$ of $G$. The set $E$ represents equations like: ::$x_1^2g_1^4x_3=1$::$x_3^2g_2x_4g_1=1$::$dots$The set $I$ represents inequations like::$g_5^\left\{-1\right\}x_3 e 1$::$dots$

By a solution in $G$ to this finite set of equations and inequations, we mean a homomorphism $f:F ightarrow G$, such that $ilde\left\{f\right\}\left(e\right)=1$ for all $ein E$ and $ilde\left\{f\right\}\left(i\right) e 1$ for all $iin I$. Where $ilde\left\{f\right\}$ is the unique homomorphism $ilde\left\{f\right\}:Fstar G ightarrow G$ that equals $f$ on $F$ and is the identity on $G$.

This formalizes the idea of substituting elements of $G$ for the variables to get true identities and inidentities. In the example the substitutions $x_1mapsto g_6, x_3mapsto g_7$ and $x_4mapsto g_8$ yield:::$g_6^2g_1^4g_7=1$::$g_7^2g_2g_8g_1=1$::$dots$::$g_5^\left\{-1\right\}g_7 e 1$::$dots$

We say the finite set of equations and inequations is consistent with $G$ if we can solve them in a "bigger" group $H$. More formally:

The equations and inequations are consistent with $G$ if there is a group$H$ and an embedding $h:G ightarrow H$ such that the finite set of equations and inequations $ilde\left\{h\right\}\left(E\right)$ and $ilde\left\{h\right\}\left(I\right)$ has a solution in $H$. Where $ilde\left\{h\right\}$ is the unique homomorphism $ilde\left\{h\right\}:Fstar G ightarrow Fstar H$ that equals $h$ on $G$ and is the identity on $F$.

Now we formally define the group $A$ to be algebraically closed if every finite set of equations and inequations that has coefficients in $A$ and is consistent with $A$ has a solution in $A$.

Known Results

It is difficult to give concrete examples of algebraically closed groups as the following results indicate:

* Every countable group can be embedded in a countable algebraically closed group.
* Every algebraically closed group is simple.
* No algebraically closed group is finitely generated.
* An algebraically closed group cannot be recursively presented.
* A finitely generated group has solvable word problem if and only if it can embedded in every algebraically closed group.

The proofs of these results are, in general very complex. However a sketch the proof that a countable group $C$ can be embedded in an algebraically closed group follows.

First we embed $C$ in a countable group $C_1$ with the property that every finite set of equations with coefficients in $C$ that is consistent in $C_1$ has a solution in $C_1$ as follows:

There are only countably many finite sets of equations and inequations with coefficients in $C$. Fix an enumeration $S_0,S_1,S_2,dots$ of them. Define groups $D_0,D_1,D_2,dots$ inductively by:

::$D_0 = C$

::

Now let:

::$C_1=cup_\left\{i=0\right\}^\left\{infty\right\}D_\left\{i\right\}$

Now iterate this construction to get a sequence of groups $C=C_0,C_1,C_2,dots$ and let:

::$A=cup_\left\{i=0\right\}^\left\{infty\right\}C_\left\{i\right\}$

Then $A$ is a countable group containing $C$. It is algebraically closed because any finite set of equations and inequations that is consistent with $A$ must have coefficients in some $C_i$ and so must have a solution in $C_\left\{i+1\right\}$.

References

* A. Macintyre: On algebraically closed groups, ann. of Math, 96, 53-97 (1972)
* B.H. Neumann: A note on algebraically closed groups. J. London Math. Soc. 27, 227-242 (1952)
* B.H. Neumann: The isomorphism problem for algebraically closed groups. In: Word Problems, pp 553-562. Amsterdam: North-Holland 1973
* W.R. Scott: Algebraically closed groups. Proc. Amer. Math. Soc. 2, 118-121 (1951)

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